How to prove ${n \choose p} \equiv\left\lfloor \frac{n}{p} \right\rfloor\pmod p$ 
Prove ${n \choose p} \equiv \left\lfloor \frac{n}{p} \right\rfloor \pmod p$

($n$ is natural and $p$ is a prime number)
How can i prove this statement?
 A: Lucas' theorem states that if we have two naturals $n,m$ and prime $p$, we can write $n$ and $m$ in base $p$:
$$n=n_0+n_1p+\cdots+n_kp^k ,\qquad m=m_0+m_1p+\cdots+m_kp^k$$
where $0\le n_i, m_i \le p-1$. Then:
$$\binom{n}{m}\equiv\prod_{i=0}^{k}\binom{n_i}{m_i}\bmod p$$
And so if $m=p$, we know that $p = 0 + 1 \cdot p$ in base $p$, so we get:
$$\binom{n}{p}\equiv\binom{n_0}{0} \binom{n_1}{1} \binom{n_2}{0} \binom{n_3}{0} \cdots\bmod p$$
Or:
$$\binom{n}{p}\equiv n_1 \bmod p$$
How do we get the digit $n_1$? It's like asking how to get the tens place digit in base $10$ from a number $1234$. You floor-divide it by the base (turn $1234$ into $123$) and then take that number mod the base ($123$ mod $10$ is $3$, which is the digit we were after).
Similarly, $n_1 = \lfloor\frac{n}{p}\rfloor \bmod p$, which means:
$$\binom{n}{p} \equiv \left\lfloor\frac{n}{p} \right\rfloor \bmod p$$
A: $\binom{n}{p} = \frac{n(n-1)(n-2)...(n-(p-1))}{p(p-1)!} \mod p$
There are exactly $p$ terms in the numerator. Let $n-r$ be divisible by $p$ where $0 \le r \le p-1$
$\binom{n}{p} = \frac{(n-r)}{p} \frac{(p-1)!}{(p-1)!}\mod p = \frac{n-r}{p} =\lfloor\frac{n}{p}\rfloor$
